Multiple-choice: sum of primes below $1000$

The sum of the first 168 positive integers is $\frac{168^2+168}{2}=14196$, which is greater than answer (a). The sum of the first 168 primes must be even greater than that.

you just have to decide between $11555$ and $76127$.

Notice that the first implies the average prime under $1000$ is $11555/168<69$. Which is clearly false.

We have to decide among $\text{(A)}$ and $\text{(B)}$. Note that the $26$th prime is $101$. This implies that if $p_{n}$ denotes the $n$ th prime, then $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\sum_{n=26}^{168} p_{n} > \sum_{n=26}^{168} 101 =101 \times 143=14443 >\text{(A)}=11555$$

The answer is thus $\text{(B)}$, $76127$. The answer can be confirmed through direct calculation or can be verfied here.